\newproblem{lay:4_6_33}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.6.33}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $A$ be any $2\times 3$ matrix such that $\mathrm{Rank}\{A\}=1$, let $\mathbf{u}$ be the first column of $A$, and suppose that $\mathbf{u}\neq 0$. Explain why
	there is a vector $\mathbf{v}\in\mathbb{R}^3$ such that $A=\mathbf{u}\mathbf{v}^T$. How could this construction be modified if the first column of $A$ were zero.
}{
  % Solution
	If $\mathrm{Rank}\{A\}=1$, then the dimension of the column space of $A$ is 1, meaning that all columns are multiples of a single vector. Without loss of generality,
	we may assume that the first column of $A$ is the basis of the column space. Then
	\begin{center}
		$\mathcal{M}_{2\times 3}\ni A=\begin{pmatrix}\mathbf{u} & a\mathbf{u} & b\mathbf{u}\end{pmatrix}$
	\end{center}
	The vector $\mathbf{v}$ proposed by the problem is
	\begin{center}
		$\mathbf{v}=\begin{pmatrix}1 \\ a \\ b \end{pmatrix}$
	\end{center}
	It can be easily verified that $A=\mathbf{u}\mathbf{v}^T$.
	
	If the first column of $A$ is zero, then the new vector $\mathbf{v}$ would be of the form
	\begin{center}
		$\mathbf{v}=\begin{pmatrix}0 \\ 1 \\ a \end{pmatrix}$
	\end{center}
	and now the basis of the column space of $A$ is given by its second column and not by its first column.
}
\useproblem{lay:4_6_33}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
